- #31
klm
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are you saying that m2= m1 v^2/r
Solve Centripetal Force for m_1 in Frictionless Table
- Thread starter klm
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SUMMARY
The discussion focuses on calculating the speed required for mass m_1 to rotate in a circle on a frictionless table while being connected to a hanging mass m_2. The key relationship established is that the centripetal force acting on m_1 must equal the weight of m_2, leading to the equation m_2g = m_1(v^2/r). Participants clarify that the tension in the string is the same for both masses and emphasize the importance of free body diagrams in understanding the forces involved. The final formula derived for speed is v = √(m_2gr/m_1).
PREREQUISITES- Understanding of Newton's laws of motion
- Knowledge of centripetal force and circular motion concepts
- Familiarity with free body diagrams (FBD)
- Basic algebra for manipulating equations
- Study the principles of circular motion and centripetal acceleration
- Learn how to construct and analyze free body diagrams
- Explore the relationship between tension and gravitational force in connected systems
- Practice solving problems involving multiple masses and forces in equilibrium
Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of problem-solving in dynamics involving circular motion and forces.
- #32
klm
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so m2g= m1 v^2/r
- #33
Kurdt
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klm said:
Almost. Remember weight is the mass multiplied by the acceleration due to gravity.
W_2 = m_2g
Can you work out v from there?
- #34
klm
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square root (m2xgxr/ m1) = v ?
- #35
Kurdt
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klm said:
Yes. Sorry I wasn't clearer earlier.
- #36
klm
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no i am sorry. i have a hard time understanding physics, and i really appreciate the help you have given me.
- #37
Kurdt
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So do you understand how this problem works now more importantly?
- #38
klm
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i think so, though i do have one question. how come normal force and weight canceled each other out in m1? does that always happen in centripetal force?
- #39
Kurdt
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Well m1 is resting on a table which means its weight is counter acted by a normal force. That just comes from Newton's 3rd law of equal and opposite reactions. Also the weight and the normal force on m1 act perpendicular to the centripetal force so they don't really come into the problem.
If I'm interpreting you wrong and you're asking why the centripetal force canceled the weight then simply because that's what the question asked you to do. So you set both equations equal and found out what speed m1 would have to travel in a circle to balance m2. Of courseif you were to set this up in a lab and you randomly spun the mass on the table it might be going too fast or too slow to balance the weight of m2.
If that doesn't answer it then I'm not sure what weight you're referring to and what normal force.
If I'm interpreting you wrong and you're asking why the centripetal force canceled the weight then simply because that's what the question asked you to do. So you set both equations equal and found out what speed m1 would have to travel in a circle to balance m2. Of courseif you were to set this up in a lab and you randomly spun the mass on the table it might be going too fast or too slow to balance the weight of m2.
If that doesn't answer it then I'm not sure what weight you're referring to and what normal force.
- #40
saket
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klm said:
No, it is NOT a rule of thumb.
I suggest you making a free body diagram (fbd) of m_1. You already have identified the forces acting on it, when seen from ground, as the weight, normal force and the tension. (ur first thread.)
In the fbd of m_1 u will see that, in this problem, in the vertical direction only weight and normal force are acting. But since m_1 is neither lifting up from the table nor breaking into the table, (note that m_1 has to move in the plane of table), acceleration of m_1 in vertical direction is zero. Using Newton's 2nd Law: m_1.g - N = m_1.(0) => N = m_1.g
We left out tension, T, in so far discussion of fbd because it was in a perpendicular to the above two forces and as u very well be knowing that component in a perpendicular direction is FCos90 = 0.
Now in the horizontal direction (ie, along the plane of table), only force is this tension. Therefore, this must provide necessary centripetal force for the m_1 to move in a circular path.
- #41
saket
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Kurdt said:
it's okay sir. i m just a new entry to this forum.. i m still learning how to handle things. go on.. u r recognized homework-helper.
- #42
klm
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thank you both very much. i understand now about the forces canceling out.